3x^2-23x+17=0

Solution for 3x^2-23x+17=0 equation:

3x^2-23x+17=0

a = 3; b = -23; c = +17;
Δ = b2-4ac
Δ = -232-4·3·17
Δ = 325
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{325}=\sqrt{25*13}=\sqrt{25}*\sqrt{13}=5\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-23)-5\sqrt{13}}{2*3}=\frac{23-5\sqrt{13}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+5\sqrt{13}}{2*3}=\frac{23+5\sqrt{13}}{6}
$